L-2: Difference between revisions

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From which you get <math> z=2 </math> in any dimension and a rough interface below <math> d=2 </math> with <math> \alpha =(2-d)/2 </math>.
From which you get <math> z=2 </math> in any dimension and a rough interface below <math> d=2 </math> with <math> \alpha =(2-d)/2 </math>.


=== Exercise L2-A: Solve Edward-Wilkinson ===
=== Exercise L2-A: Solve Edwards-Wilkinson ===


For simplicity, consider a 1-dimensional line of size L with periodic boundary conditions. It is useful to introduce the Fourier modes:
For simplicity, consider a 1-dimensional line of size L with periodic boundary conditions. It is useful to introduce the Fourier modes:
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* Compute <math> \langle \cal{E}_q (t)\rangle  </math> which describes how the noise injects the energy on the different modes. Comment about  equipartition and the dynamical exponent   
* Compute <math> \langle \cal{E}_q (t)\rangle  </math> which describes how the noise injects the energy on the different modes. Comment about  equipartition and the dynamical exponent   
* Compute the width  <math> \langle h(x,t)^2\rangle = \sum_q \langle h_q(t)h_{-q}(t) \rangle  </math>. Comment about the roughness and the short times growth.
* Compute the width  <math> \langle h(x,t)^2\rangle = \sum_q \langle h_q(t)h_{-q}(t) \rangle  </math>. Comment about the roughness and the short times growth.
= KPZ equation and interface growth=
= KPZ equation and interface growth=



Revision as of 14:45, 8 January 2024

Goal: The physical properties of many materials are controlled by the interfaces embedded in it. This is the case of the dislocations in a crystal, the domain walls in a ferromagnet or the vortices in a supercoductors. In the next lecture we will discuss how impurities affect the behviour of these interfaces. Today we focus on thermal fluctuations and introduce two important equations for the interface dynamics: the Edwards Wilkinson (EW) and the Kardar Parisi Zhang (KPZ) equations.

Edwards Wilkinson: an interface at equilibrium:

Consider domain wall h(r,t) fluctuating at equilibrium at the temparature T. Here t is time, r defines the d-dimensional coordinate of the interface and h is the scalar height field. Hence, the domain wall separating two phases in a film has d=1,r, in a solid instead d=2,r𝟐.

Two assumptions are done:

  • Overhangs, pinch-off are neglected, so that h(r,t) is a scalar univalued function.
  • The dynamics is overdamped, so that we can neglect the inertial term.

Derivation

The Langevin equation of motion is

th(r,t)=μδEpotδh(r,t)+η(r,t)

The first term δEpot/δh(r,t) is the elastic force trying to smooth the interface, the mobility μ is inversily proportional to the viscosity. The second term is the Langevin Gaussian noise defined by the correlations

η(r,t)=0,η(r,t)η(r,t)=2Dδd(rr)δ(tt)

The symbol indicates the average over the thermal noise. The diffusion constant is fixed by the Eistein relation (fluctuation-dissipation theorem):

D=μKBT

The potential energy of surface tension can be expanded at the lowest order in the gradient:

Epot=σddr1+(h)2const.+σ2ddr(h)2

Setting μ=1,σ=1/2 we have the Edwards Wilkinson equation:

th(r,t)=122h(r,t)+η(r,t)

Scaling Invariance

The equation enjoys of a continuous symmetry because h(r,t) and h(r,t)+c cannot be distinguished. This is a conndition os scale invariance:

h(br,bzt)inlawbαh(r,t)

Here z,α are the dynamic and the roughness exponent rispectively. From dimensional analysis

bαzth(r,t)=bα22h(r,t)+bd/2z/2η(r,t)

From which you get z=2 in any dimension and a rough interface below d=2 with α=(2d)/2.

Exercise L2-A: Solve Edwards-Wilkinson

For simplicity, consider a 1-dimensional line of size L with periodic boundary conditions. It is useful to introduce the Fourier modes:

h^q(t)=1L0Leiqrh(r,t),h(r,t)=qeiqrh^q(t)

Here q=2πn/L,n=,1,0,1, and recall 0Ldreiqr=Lδq,0.

  • Show that the EW equation writes
th^q(t)=μσq2h^q(t)+ηq(t),withηq1(t)ηq2(t)=DLδq1,q2δ(tt)

The solution of this first order linear equation writes

h^q(t)=h^q(0)+0tdseμσq2sηq(s)

Assume that the interface is initialy flat, namely h^q(0)=0. Note that Epot(t)=qq(t)=Lσ2qq2hq(t)hq(t)

  • Compute E𝓆(𝓉) which describes how the noise injects the energy on the different modes. Comment about equipartition and the dynamical exponent
  • Compute the width h(x,t)2=qhq(t)hq(t). Comment about the roughness and the short times growth.

KPZ equation and interface growth

Consider a domain wall in presence of a positive magnetic field. At variance with the previous case the ferromagnetic domain aligned with the field will expand while the other will shrink. The motion of the interface describes now the growth of the stable domain, an out-of-equilibrium process.

Derivation

To derive the correct equation of a growing interface the key point is to realize that the growth occurs locally along the normal to the interface (see figure).

Let us call v the velocity of the interface. Consider a point of the interface h(r,t), its tangent is rh(r,t)=tan(θ). To evaluate the increment $\delta h(r,t)$ use the Pitagora theorem:

δh(r,t)=(vdt)2+(vdttan(θ))2vdt+vdt2(tan(θ))2vdt+vdt2(rh(r,t))2

Hence, in generic dimension, the KPZ equation is

th(r,t)=ν2h(r,t)+λ2(h)2+η(r,t)

Scaling Invariance

The symmetry h and h+c still holds so that scale invariance is still expected. However the non-linearity originate an anomalous dimension and z,α cannot be determined by simple dimensional analysis. Moreover, the previous scaling relation does not holds because the non-linear term is non-conservative:

dr(h)2>0so thatdrth(r,t)drη(r,t)

An important symmetry

Let us remark that if h(r,t) is a solution of KPZ,also h~(rλv0t,t)+v0r(v02/2)t is a solution of KPZ.

You can check it, and you will obtain an equation with the statistically equivalent noise η~(r,t)=η(rλv0t,t). The symmetry relies on two properties:

  • The noise η(r,t) is delta correlated in time
  • Only sticked together the two terms th(r,t) and λ2(rh(r,t))2 enjoy the symmetry. Hence, under the rescaling

bαz(th(r,t)bα+z2λ2(rh(r,t))2)

the second term should be b-independent. This provides a new and exact scaling relation

z+α=2

Stochastic Burgers and Galilean invariance

Let's discuss for simplicity the 1-dimensional case and perform the following change of variable

v(r,t)=λrh(r,t)

You obtain the Brugers equation

tv(r,t)+vrv=

Check that the important symmetry of the KPZ equation is nothing but the Galilean symmetry of the Burgers equation.

The d=1 case